How to understand noise in digital camera photos? A common believe is that it is related to ISO. The higher the ISO, the stronger the noise. However, some also said ISO has nothing to do with noise. To be accurate, the only valid statement is that noise is a built-in property of the image sensor. It is complicated to tell how would that translate into the noise you see in the image.

The digital image sensor is to be activated by photons and respond with voltage output. Reading the dataset of a digital image sensor, some parameters would be related to this function:

  • Sensitivity: In unit of volt per lux-second
  • Dark current: In unit of volt per second

The sensitivity depends on multiple factors, including the wavelength of light, temperature, and light intensity. Usually the sensitivity is nonlinear to these parameters. The dark current may also depends on temperature. Ideally, the sensor should be a function that translate the light intensity into voltage. In reality, there is a non-zero noise added.

Therefore, mathematically, we may represent the response of the image sensor with:

\[I = \int_0^T g \big( L_A(t) + \epsilon(t)\big) dt\]

where $\epsilon(t)$ represents the noise function and $L_A(t)$ the sensor function. In a simplified model, $\epsilon(t)$ is a small random quantity and $L_A(t)$ is a function depends on the aperture of the camera and light intensity of the subject. Both are a function of time. The constant $g$ is the gain, which is controlled by ISO. The time $T$ is the reciprocal of the shutter speed. $I$ is the pixel value as produced from the camera (before clipping due to overexposure).

The image sensor gives you electrical signal representation of the light. ISO is implemented as the amplifier to the signal. The signal is produced by accumulating photons on the sensor in which will charge up the voltage output. This is a process that takes time.

By looking at the integral above, the noise as perceived is determined by the signal-to-noise ratio $L_A(t)/\epsilon(t)$. Really, ISO plays no role here. Neither do the length of exposure $T$. To reduce noise, the stronger incident light is all you need. The reason for the misconception is at low-light situation, to make $I$ not underexposed, the natural operation is to increase $g$ or $T$ while the aperture was limited by the lens so $L_A(t)$ is not adjustable.